We have the real positive(only) numbers space of dimension n.We have also a relation of two members(sorry),continuous,absoletuly monotonic(only >),cursive,in this space.
Also x,y are arrays and x>y in this space and 0<a<1.
How can we proove that a*x+(1-a)*y>y ?. (*: multiplication)
(G. Aliprantis, D. Brown and O. Burhinshaw, Existence and Optimality of Competitive Equilibrium, Springer-Verlag, 1990.)
sorry for my english.
Last edited by razor (2006-03-05 10:59:29)
ax + (1-a)y > ay + (1-a)y = y(a + 1 - a) = y
ax + (1-a)y > y
All that needed was x > y.
Last edited by Ricky (2006-03-05 12:56:13)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."